420
INSTRUCTOR’S SOLUTIONS MANUAL
energy as
Ea = RT 2
!
d ln k
dT
"
[22.30]
This definition reduces to the earlier one (as the slope of a straight line) for a temperature-independent
activation energy. However, this latter definition is more general, because it allows Ea to be obtained
from the slope (at the temperature of interest) of a plot of ln k against 1/T even if the Arrhenius plot is
not a straight line. Non-Arrhenius behavior is sometimes a sign that quantum mechanical tunnelling is
playing a significant role in the reaction.
D22.10
The expression k = ka kb [A]/(kb + ka! [A]) for the effective rate constant of a unimolecular reaction
ka
A → P is based on the validity of the assumption of the existence of the pre-equilibrium A+A ! A∗ +A.
ka!
This can be a good assumption if both ka and ka! are much larger than kb . The expression for the effective
rate-constant, k, can be rearranged to
k!
1
1
= a +
k
ka k b
ka [A]
Hence, a test of the theory is to plot 1/k against 1/[A], and to expect a straight line. Another test is based
on the prediction from the Lindemann–Hinshelwood mechanism that as the concentration (and therefore
the partial pressure) of A is reduced, the reaction should switch to overall second order kinetics. Whereas
the mechanism agrees in general with the switch in order of unimolecular reactions, it does not agree in
detail. A typical graph of 1/k against 1/[A] has a pronounced curvature, corresponding to a larger value
of k (a smaller value of 1/k) at high pressures (low 1/[A]) than would be expected by extrapolation of
the reasonably linear low pressure (high 1/[A]) data.
Solutions to exercises
E22.1(a)
v=
1 d[J]
[22.3b]
νJ dt
so
d[J]
= νJ v.
dt
The reaction has the form
0 = 3C + D − A − 2B.
Rate of formation of C = 3v = 3.0 mol dm−3 s−1 .
Rate of formation of D = v = 1.0 mol dm−3 s−1 .
Rate of formation of A = v = 1.0 mol dm−3 s−1 .
Rate of formation of B = 2v = 2.0 mol dm−3 s−1 .
E22.2(a)
v=
1 d[J]
1 d[C]
1
=
= × (1.0 mol dm−3 s−1 ) = 0.50 mol dm−3 s−1 [22.3b].
νJ dt
2 dt
2
Rate of formation of D = 3v = 1.5 mol dm−3 s−1 .
Rate of consumption of A = 2v = 1.0 mol dm−3 s−1 .
Rate of consumption of B = v = 0.50 mol dm−3 s−1 .
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THE RATES OF CHEMICAL REACTIONS
E22.3(a)
421
The rate is expressed in mol dm−3 s−1 ; therefore
mol dm−3 s−1 = [k] × (mol dm−3 ) × (mol dm−3 ),
where [k] denotes units of k, requires the units to be dm3 mol−1 s−1 .
(a) Rate of formation of A = v = k[A][B] .
(b) Rate of consumption of C = 3v = 3k[A][B] .
E22.4(a)
d[C]
= k[A][B][C],
dt
the rate of reaction is (eqn 22.3b)
Given
v=
1 d[C]
1 d[J]
=
=
νJ dt
2 dt
1
2 k[A][B][C]
.
The units of k, [k], must satisfy
mol dm−3 s−1 = [k] × (mol dm−3 ) × (mol dm−3 ) × (mol dm−3 ).
Therefore, [k] = dm6 mol−2 s−1 .
E22.5(a)
The rate law is
v = k[A]a ∝ paA = {pA,0 (1 − f )}a ,
where f is the fraction reacted. That is, concentration and partial pressure are proportional to each other.
Thus we can write
!
"
paA,1
v1
1 − f1 a
= a =
.
v2
pA,2
1 − f2
Taking logarithms
! "
!
"
v1
1 − f1
log
= a log
v2
1 − f2
$
#
log vA,1 /vA,2
log (1.07/0.76)
=
= 1.99.
so a =
log[(1 − f1 )/(1 − f2 )]
log (0.95/0.80)
Hence, the reaction is second-order .
COMMENT. Knowledge of the initial pressure is not required for the solution to this exercise. The ratio of
pressures was computed using fractions of the initial pressure.
E22.6(a)
Table 22.3 gives a general expression for the half-life of a reaction of the type A → P for orders greater
than 1:
t1/2 =
2n−1 − 1
(n − 1)k[A]0n−1
∝ [A]01−n ∝ p01−n
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INSTRUCTOR’S SOLUTIONS MANUAL
where the proportionality constants may be functions of the reaction order, the rate constant, or even the
% &1−n % &n−1
t1/2 (p0,1 )
p0,1
p
temperature, but not of the concentration. So t1/2
= p0,2
.
(p0,2 ) = p0,2
0,1
"
!
"
!
t1/2 (p0,1 )
p0,2
= (n − 1) log
Hence, log
t1/2 (p0,2 )
p0,1
or (n − 1) =
log (410/880)
= 0.999 ≈ 1.
log (169/363)
Therefore, n = 2 in agreement with the result of Exercise 22.5(a).
E22.7(a)
v = k[N2 O5 ].
2N2 O5 → 4NO2 + O2
Therefore, rate of consumption of N2 O5 = 2v = 2k[N2 O5 ].
d[N2 O5 ]
= −2k[N2 O5 ]
dt
so [N2 O5 ] = [N2 O5 ]0 e−2kt .
Solve this for t:
t=
1
[N2 O5 ]0
.
ln
[N2 O5 ]
2k
Therefore, the half-life is:
t1/2 =
1
ln 2
= 1.03 × 104 s .
ln 2 =
2k
(2) × (3.38 × 10−5 s−1 )
Since the partial pressure of N2 O5 is proportional to its concentration,
p(N2 O5 ) = p0 (N2 O5 )e−2kt .
(a) p(N2 O5 ) = (500 Torr) × (e−(6.76×10
(b) p(N2 O5 ) =
−5 /s)×(10 s)
) = 4.997 Torr .
−5
(500 Torr) × (e−(6.76×10 /s)×(600 s) )
= 480 Torr .
COMMENT. The half-life formula in eqn 22.13 is based on a rate constant for the rate of change of the
reactant, that is, based on the assumption that
−
d[A]
= k[A].
dt
Our expression for the rate of consumption has 2k instead of k, and our expression for t1/2 does likewise.
E22.8(a)
The integrated rate law is
kt =
1
ln
[B]0 − [A]0
'!
[B]
[B]0
"(!
[A]
[A]0
")
[22.19].
(a) The stoichiometry of the reaction requires that, when
"[A] = (0.020 − 0.050) mol dm−3 = −0.030 mol dm−3 ,
then "[B] = −0.030 mol dm−3 as well.
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INSTRUCTOR’S SOLUTIONS MANUAL
Solving for x yields, after some rearranging,
x=
=
(a)
[A]0 [B]0 {ek([B]0 −[A]0 )t − 1}
(0.050) × (0.100 mol dm−3 ) × {e(0.100−0.050)×0.11×t/s − 1}
=
([B]
−[A]
)kt
0
(0.100) × {e(0.100−0.050)×0.11×t/s } − 0.050
[B]0 e 0
− [A]0
(0.100 mol dm−3 ) × (e5.5×10
2e
5.5×10−3 t/s
−3 t/s
−1
− 1)
.
(0.100 mol dm−3 ) × (e0.055 − 1)
= 5.1 × 10−3 mol dm−3
2e0.055 − 1
which implies that [NaOH] = (0.050 − 0.0051) mol dm−3 = 0.045 mol dm−3 and
x=
[CH3 COOC2 H5 ] = (0.100 − 0.0051) mol dm−3 = 0.095 mol dm−3
(b)
(0.100 mol dm−3 ) × (e3.3 − 1)
= 0.049 mol dm−3 .
2e3.3 − 1
Hence, [NaOH] = (0.050 − 0.049) mol dm−3 = 0.001 mol dm−3 and
x=
[CH3 COOC2 H5 ] = (0.100 − 0.049) mol dm−3 = 0.051 mol dm−3 .
E22.11(a) The rate of consumption of A is
−
d[A]
= 2v = 2k[A]2
dt
[νA = −2]
1
1
= 2kt [22.15b with k replaced by 2k].
−
[A] [A]0
!
"
1
1
1
,
Therefore, t =
−
2k [A] [A]0
"
!
" !
1
1
1
−
t=
×
0.011 mol dm−3
0.260 mol dm−3
(2) × (3.50 × 10−4 dm3 mol−1 s−1 )
which integrates to
= 1.24 × 105 s .
E22.12(a) The rate of change of [A] is
d[A]
= −k[A]n .
dt
# [A]
# t
d[A]
Hence,
=
−k
dt = −kt.
n
[A]0 [A]
0
%
" $
!
1
1
1
×
−
.
Therefore, kt =
n−1
[A]n−1
[A]0n−1
At t = t1/2 , [A] = [A]0 /2
% !
%
!
" $ n−1
" $
2
1
1
1
2n−1 − 1
×
×
−
=
[as in Table 22.3].
and kt1/2 =
n−1
n−1
[A]0n−1
[A]0n−1
[A]0n−1
Hence, t1/2 ∝
1
[A]0n−1
.
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THE RATES OF CHEMICAL REACTIONS
425
E22.13(a) The reaction for which pKa is 9.25 is
+
NH+
4 (aq) + H2 O(l) ! NH3 (aq) + H3 O (aq).
The reaction whose rate constant we are to find is the forward reaction in the following equilibrium
k
−
NH3 (aq) + H2 O(l) ! NH+
4 (aq) + OH (aq)
k$
Kb .
The equilibria are related by
pKb = pKw − pKa = 14.00 − 9.25 = 4.75.
Therefore, Kb =
k
= 10−4.75 = 1.78 × 10−5 mol dm−3
k$
and k = Kb k $ = (1.78 × 10−5 mol dm−3 ) × (4.0 × 1010 dm3 mol−1 s−1 ) = 7.1 × 105 s−1 .
Now proceed as in Example 22.4:
1
−
= k + k $ ([NH+
4 ] + [OH ])
τ
= k + 2k $ (Kb [NH3 ])1/2
−
1/2
[[NH+
]
4 ] = [OH ] = (Kb [NH3 ])
= 7.1 × 105 s−1 + 2 × (4.0 × 1010 dm3 mol−1 s−1 )
× [(1.78 × 10−5 ) × (0.15)]1/2 mol dm−3
= 1.31 × 108 s−1 .
Hence τ = 7.61 ns .
COMMENT. The rate constant k corresponds to the pseudo first-order protonation of NH3 in excess water
and hence has the units s−1 . Therefore, Kb in this problem must be assigned the units mol dm−3 to obtain
proper cancellation of units.
E22.14(a) Call the rate constant k at temperature T , and the rate constant k $ at temperature T $ :
Ea
Ea
, ln k $ = ln A −
[22.29].
RT
RT $
&
'
&
'
R ln k $ /k
(8.314 J K −1 mol−1 ) × ln 1.38 × 10−2 /2.80 × 10−3
=
Hence, Ea =
(1/303 K) − (1/323 K)
(1/T ) − (1/T $ )
ln k = ln A −
= 64.9 kJ mol−1 .
For A, we rearrange eqn 22.31:
A = k × eEa /RT = (2.80 × 10−3 mol dm−3 s−1 ) × e64.9×10
3 /(8.314×303)
= 4.32 × 108 mol dm−3 s−1 .
−D or D−
−H bond is involved in the rate-determining step, use eqn 22.53:
E22.15(a) If cleavage of a C−
−D)
k(C−
= e−λ ,
−H)
k(C−
(
" )
!
−H)
hcν̃(C−
µCH 1/2
.
λ=
1−
2kB T
µCD
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THE RATES OF CHEMICAL REACTIONS
433
COMMENT. The data define a good straight line, as the correlation coefficient R2 = 0.996 shows. That
straight line appears to go through the origin, but the best-fit equation gives a small non-zero y-intercept.
Inspection of the plot shows that several of the data points lie about as far from the fit line as the y-intercept
does from zero. This suggests that y-intercept has a fairly high relative uncertainty, and so do the rate
constants.
P22.16
Apply the equation derived in P22.5 to the rate constant data in pairs
!
"
−R ln k/k "
Ea =
((1/T ) − (1/T " ))
T /K
300.3
300.3
341.2
T " /K
341.2
392.2
392.2
10−7 k/(dm3 mol−1 s−1 )
10−7
k " /(dm3
1.44
mol−1 s−1 )
3.03
Ea /(kJ mol−1 )
15.5
1.44
3.03
6.9
6.9
16.7
18.0
The mean is 16.7 kJ mol−1 . Compute A from each rate constant, using the mean Ea and A = keEa /RT .
300.3
T /K
10−7 k/(dm3 mol−1 s−1 )
341.2
1.44
392.2
3.03
6.9
Ea /RT
6.69
5.89
5.12
10−10 A/(dm3 mol−1 s−1 )
1.16
1.10
1.16
The mean is 1.14 × 1010 dm3 mol−1 s−1
P22.18
The relation between the equilibrium constant and the rate constants is obtained from
k
k"
#
$
#
$ # $
# "
$
−!r H $
Ea − Ea
k
!r S $
A
exp
So K = " = exp
exp
=
k
RT
R
A"
RT
!r G $ = −RT ln K = !r H $ − T !r S $
K=
and
Setting the temperature-dependent parts equal yields
!r H $ = Ea − Ea" = [−4.2 − (53.3)] kJ mol−1 = −57.5 kJ mol−1
Setting the temperature-independent parts equal yields
#
$
!r S $
R
$
=
#
A
A"
so !r S $ = R ln
#
A
A"
$
= (8.3145 J K−1 mol−1 ) ln
exp
#
1.0 × 109
1.4 × 1011
$
= −41.1 J K−1 mol−1
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INSTRUCTOR’S SOLUTIONS MANUAL
The thermodynamic quantities of the reaction are related to standard molar quantities
!r H $ = !f H $ (C2 H6 ) + !f H $ (Br) − !f H $ (C2 H5 ) − !f H $ (HBr)
so !f H $ (C2 H5 ) = !f H $ (C2 H6 ) + !f H $ (Br) − !f H $ (HBr) − !r H $
and !f H $ (C2 H5 ) = [(−84.68) + 111.88 − (−36.40) − (−57.5)] kJ mol−1 = 121.2 kJ mol−1
Similarly
Sm$ (C2 H5 ) = [229.60 + 175.02 − 198.70 − (−41.1)] J mol−1 K−1 = 247.0 J K−1 mol−1
Finally
!f G $ (C2 H5 ) = [−32.82 + 82.396 − (−53.45)] kJ mol−1 − !r G $
= 103.03 kJ mol−1 − !r G $
but
!r G $ = !r H $ − T !r S $ = −57.5 kJ mol−1 − (298 K) × (−41.1 × 10−3 kJ K−1 mol−1 )
= −45.3 kJ mol−1
so !f G $ (C2 H5 ) = [103.03 − (−45.3)] kJ mol−1 = 148.3 kJ mol−1
Solutions to theoretical problems
P22.20
We assume a pre-equilibrium (as the initial step is fast), and write
K=
[A]2
,
[A2 ]
implying that [A] = K 1/2 [A2 ]1/2
The rate-determining step then gives
ν=
d[P]
= k2 [A][B] = k2 K 1/2 [A2 ]1/2 [B] = keff [A2 ]1/2 [B]
dt
where keff = k2 K 1/2 .
P22.22
d[P]
= k[A][B]
dt
Let the initial concentrations be [A]0 = A0 , [B]0 = B0 , and [P]0 = 0. Then, when P is formed in
concentration x, the concentration of A changes to A0 − 2x and that of B changes to B0 − 3x. Therefore
ν=
d[P]
dx
=
= k(A0 − 2x)(B0 − 3x)
dt
dt
with
x = 0 at t = 0.
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THE RATES OF CHEMICAL REACTIONS
%
t
%
435
x
dx
(A0 − 2x) × (B0 − 3x)
$ #
$
% x#
6
1
1
=
×
−
dx
2B0 − 3A0
3(A0 − 2x) 2(B0 − 3x)
0
$
$ #% x
#
% x
dx
dx
−1
×
−
=
2B0 − 3A0
0 x − (1/3)B0
0 x − (1/2)A0
()
(
'
&
'
#
$
x − 13 B0
x − 21 A0
−1
kt =
− ln
× ln
(2B0 − 3A0 )
− 21 A0
− 13 B0
$ #
$
#
−1
(2x − A0 )B0
=
ln
2B0 − 3A0
A0 (3x − B0 )
#
$ #
$
(2x − A0 )B0
1
ln
=
(3A0 − 2B0 )
A0 (3x − B0 )
P22.24
k dt =
0
0
The rate equations are
d[A]
= −ka [A] + ka" [B]
dt
d[B]
= ka [A] − ka" [B] − kb [B] + kb" [C]
dt
d[C]
= kb [B] − kb" [C]
dt
These equations are a set of coupled differential equations and, though it is not immediately apparent,
they do admit of an analytical general solution. However, we are looking for specific circumstances under
which the mechanism reduces to the second form given. Since the reaction involves an intermediate, let
us explore the result of applying the steady-state approximation to it. Then
d[B]
= ka [A] − ka" [B] − kb [B] + kb" [C] = 0
dt
and [B] =
Therefore,
ka [A] + kb" [C]
ka" + kb
k" k"
d[A]
k a kb
=− "
[A] + " a b [C]
dt
ka + kb
ka + kb
This rate expression may be compared to that given in the text [Section 22.4] for the mechanism
)
&
k
A!B
k"
Hence, keff =
keff
here A ! C
"
keff
k a kb
+ kb
ka"
The solutions are [A] =
" =
keff
'
ka" kb"
+ kb
ka"
"
" + k e−(keff +keff )t
keff
eff
" +k
keff
eff
(
× [A]0 [22.23]
and [C] = [A]0 − [A]
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